Each integer from -3 to 5 inclusive is placed in the diagram, with one number going into each box. If the sum of the numbers in each row is the same, what is that sum? (the diagram will be a screenshot) A. 0 b. 3 c. 5 d.6

Our solution is option b, or in other words 3.

Step-by-step explanation:

This is more of a " guess and try " question. Say that 3 is the solution. If that is so, let us prove it, by arranging the values such that their addition is equivalent to 3 in each row.

Take - 3 as the first value in the first row. What other combinations can go with this, but still meet our requirement of a total of 3? The other two numbers must have a sum of 6, so let's list all possibilities -

{ 5 and 1 }

{ 4 and 2 }

Let us try 5 and 1. We should receive the following table so far -

| - 3 | 5 | 1 |

|   ?  | ? | ? |

|   ?  | ? | ? | Now for this second row let - 2 be the starting value. The only possible values in the row should have a sum of 5 to have a total of 3 -

{ 3 and 2 }

This is the only possibility we have here. Plug it into our table,

| - 3 | 5 | 1 |

| - 2 | 3 | 2 |

|   ?  | ? | ? | Let's begin with the value - 1 for the third row. The only remaining values that could fill this row are 0 and 4. { - 1, 0, and 4 } add to 3, and hence the sum of the number in each row is a constant, 3. It has been proved.

| - 3 | 5 | 1 |

| - 2 | 3 | 2 | - This is our table.

| - 1  | 0 | 4 |

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.